Android - send parameter to php script to use it in query

Question

In my app data is fetched from a database. For that a variable is passed to a php script and should be used in the query later. I tried it using $var = $_POST['name'] in the query, but it seems like the passed variable is empty. To test it, I inserted the variable into a table and used the isset() method as well. The first part works as it should, the variable is set and inserted, but in the second part with the query to fetch the data it doesn’t seem to be set since the variable returns the value „a“. Why is the value not set and not taken from above?

The php code:

<?php

include Connection.php;



$conn = new mysqli($servername, $username, $password, $dbname);


if(isset($_POST['name'])){

 $query = "INSERT INTO TestTable (name) VALUES ('$_POST[name]')";


 if(mysqli_query($conn,$query)){

    echo 'Data Submit Successfully';

 }
 else{

    echo 'Try Again';

 }
}



if(!isset($_POST['name'])){
    $var = "a";
} 

 $query = "SELECT * FROM TestTable WHERE name = '$var'";

    $result = mysqli_query($conn, $query);

    while($row = mysqli_fetch_assoc($result)) {
            $array[] = $row;
        }



    header('Content-Type:Application/json');
    echo json_encode($array);
    mysqli_close($conn);

?>

change "INSERT INTO TestTable (name) VALUES ('$_POST[name]')" to "INSERT INTO TestTable (name) VALUES ($_POST['name'])" — Umer Farooq
– Umer Farooq, Commented Apr 30, 2018 at 10:04
@UmerFarooq Sorry for unprecise questioning, I tried to improve it. The problem I have is in the fetching part, my bad! — Mike Kng
– Mike Kng, Commented Apr 30, 2018 at 10:06
have you tried query without variable i mean "SELECT * FROM TestTable WHERE name = 'a'" — Umer Farooq
– Umer Farooq, Commented Apr 30, 2018 at 10:15
@UmerFarooq yes I tried that and that works as it should do as well! — Mike Kng
– Mike Kng, Commented Apr 30, 2018 at 20:41

Alexey Usachov · Accepted Answer · 2018-04-30 10:23:16Z

2

Your query string is incorrect, fix by this:

//...
$query = "INSERT INTO TestTable (name) VALUES ('".$_POST['name']."')";
//...

To fix SELECT query you should initialize $var in case isset($_POST['name']) returns true

//...
$var = "a";
if(isset($_POST['name'])){
    $var = $_POST['name'];
} 

$query = "SELECT * FROM TestTable WHERE name = '$var'";

edited Apr 30, 2018 at 10:23

answered Apr 30, 2018 at 10:02

Alexey Usachov

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2 Comments

Mike Kng Over a year ago

The inserting works this way, is it wrong or unconventional?

Alexey Usachov Over a year ago

@MikeKng string can't parse array with key expression, you need to do it with a help of string concatenation or with a help of additional variable. ('".$_POST['name']."')"; or $var = $_POST['name']; $query = "INSERT INTO TestTable (name) VALUES ('$var')";

Umer Farooq · Accepted Answer · 2018-04-30 10:19:03Z

2

In the code

if(!isset($_POST['name'])){
    $var = "a";
} 

 $query = "SELECT * FROM TestTable WHERE name = '$var'";

change this to

$var = "";

if(!isset($_POST['name'])){
    $var = "a";
} 

 $query = "SELECT * FROM TestTable WHERE name = '$var'";

answered Apr 30, 2018 at 10:19

Umer Farooq

7921 gold badge9 silver badges17 bronze badges

9 Comments

Mike Kng Over a year ago

Isn‘t $var just empty then?

Umer Farooq Over a year ago

No..when !isset($_POST['name']) gets true, its value will set to 'a'. $var must be declared outside the scope of if block so that you can access it out side the scope of if block.

Umer Farooq Over a year ago

Since your select query is outside the if block therefore $var must be declared outside the scope of if block otherwise you can't access $var in your query

Mike Kng Over a year ago

I understand, but $_POST['name'] is used above as well, why does the isset() return that it‘s empty?

Umer Farooq Over a year ago

have you tried debugging the script by calling var_dump($_POST) before calling isset(). ?

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